Subset of $ell^2$ with distance property












8












$begingroup$


I'd been trying the following problem:




Prove that exists an infinite set $Asubset B(0,1)$ such that $|x-y|_2>sqrt{2}$ for all $x,y in A$.




My ideas always end in points with distance at most $sqrt{2}$.



I need hints, please.






functional-analysis hilbert-spaces lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is there something missing in the question? If $A$ contains only two points $(1,0,0, ldots)$ and $(-1, 0, 0, ldots)$, does this not work?
    $endgroup$
    – angryavian
    Dec 5 '18 at 20:40










  • $begingroup$
    My mistake, A must be infinite.
    $endgroup$
    – Ángela Flores
    Dec 5 '18 at 20:40










  • $begingroup$
    @AlexR. $l^2$ refers to the space of sequences $(a_1, a_2, ldots)$ such that $sum_n a_n^2 < infty$.
    $endgroup$
    – angryavian
    Dec 5 '18 at 20:47










  • $begingroup$
    Nitpick: You might want to add $x ne y$ in the formulation of your question. Otherwise, it is not possible (obviously).
    $endgroup$
    – gerw
    Dec 6 '18 at 7:52










  • $begingroup$
    An interesting spinoff: How large can $$inf_{x,y in A, x ne y} |x-y|$$ be? Can this be larger than $sqrt{2}$?
    $endgroup$
    – gerw
    Dec 6 '18 at 8:04


















8












$begingroup$


I'd been trying the following problem:




Prove that exists an infinite set $Asubset B(0,1)$ such that $|x-y|_2>sqrt{2}$ for all $x,y in A$.




My ideas always end in points with distance at most $sqrt{2}$.



I need hints, please.






functional-analysis hilbert-spaces lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is there something missing in the question? If $A$ contains only two points $(1,0,0, ldots)$ and $(-1, 0, 0, ldots)$, does this not work?
    $endgroup$
    – angryavian
    Dec 5 '18 at 20:40










  • $begingroup$
    My mistake, A must be infinite.
    $endgroup$
    – Ángela Flores
    Dec 5 '18 at 20:40










  • $begingroup$
    @AlexR. $l^2$ refers to the space of sequences $(a_1, a_2, ldots)$ such that $sum_n a_n^2 < infty$.
    $endgroup$
    – angryavian
    Dec 5 '18 at 20:47










  • $begingroup$
    Nitpick: You might want to add $x ne y$ in the formulation of your question. Otherwise, it is not possible (obviously).
    $endgroup$
    – gerw
    Dec 6 '18 at 7:52










  • $begingroup$
    An interesting spinoff: How large can $$inf_{x,y in A, x ne y} |x-y|$$ be? Can this be larger than $sqrt{2}$?
    $endgroup$
    – gerw
    Dec 6 '18 at 8:04
















8












8








8





$begingroup$


I'd been trying the following problem:




Prove that exists an infinite set $Asubset B(0,1)$ such that $|x-y|_2>sqrt{2}$ for all $x,y in A$.




My ideas always end in points with distance at most $sqrt{2}$.



I need hints, please.






functional-analysis hilbert-spaces lp-spaces







share|cite|improve this question











$endgroup$




I'd been trying the following problem:




Prove that exists an infinite set $Asubset B(0,1)$ such that $|x-y|_2>sqrt{2}$ for all $x,y in A$.




My ideas always end in points with distance at most $sqrt{2}$.



I need hints, please.






functional-analysis hilbert-spaces lp-spaces




functional-analysis hilbert-spaces lp-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 22:43









Xander Henderson

14.3k103554




14.3k103554










asked Dec 5 '18 at 20:36









Ángela FloresÁngela Flores

26518




26518












  • $begingroup$
    Is there something missing in the question? If $A$ contains only two points $(1,0,0, ldots)$ and $(-1, 0, 0, ldots)$, does this not work?
    $endgroup$
    – angryavian
    Dec 5 '18 at 20:40










  • $begingroup$
    My mistake, A must be infinite.
    $endgroup$
    – Ángela Flores
    Dec 5 '18 at 20:40










  • $begingroup$
    @AlexR. $l^2$ refers to the space of sequences $(a_1, a_2, ldots)$ such that $sum_n a_n^2 < infty$.
    $endgroup$
    – angryavian
    Dec 5 '18 at 20:47










  • $begingroup$
    Nitpick: You might want to add $x ne y$ in the formulation of your question. Otherwise, it is not possible (obviously).
    $endgroup$
    – gerw
    Dec 6 '18 at 7:52










  • $begingroup$
    An interesting spinoff: How large can $$inf_{x,y in A, x ne y} |x-y|$$ be? Can this be larger than $sqrt{2}$?
    $endgroup$
    – gerw
    Dec 6 '18 at 8:04




















  • $begingroup$
    Is there something missing in the question? If $A$ contains only two points $(1,0,0, ldots)$ and $(-1, 0, 0, ldots)$, does this not work?
    $endgroup$
    – angryavian
    Dec 5 '18 at 20:40










  • $begingroup$
    My mistake, A must be infinite.
    $endgroup$
    – Ángela Flores
    Dec 5 '18 at 20:40










  • $begingroup$
    @AlexR. $l^2$ refers to the space of sequences $(a_1, a_2, ldots)$ such that $sum_n a_n^2 < infty$.
    $endgroup$
    – angryavian
    Dec 5 '18 at 20:47










  • $begingroup$
    Nitpick: You might want to add $x ne y$ in the formulation of your question. Otherwise, it is not possible (obviously).
    $endgroup$
    – gerw
    Dec 6 '18 at 7:52










  • $begingroup$
    An interesting spinoff: How large can $$inf_{x,y in A, x ne y} |x-y|$$ be? Can this be larger than $sqrt{2}$?
    $endgroup$
    – gerw
    Dec 6 '18 at 8:04


















$begingroup$
Is there something missing in the question? If $A$ contains only two points $(1,0,0, ldots)$ and $(-1, 0, 0, ldots)$, does this not work?
$endgroup$
– angryavian
Dec 5 '18 at 20:40




$begingroup$
Is there something missing in the question? If $A$ contains only two points $(1,0,0, ldots)$ and $(-1, 0, 0, ldots)$, does this not work?
$endgroup$
– angryavian
Dec 5 '18 at 20:40












$begingroup$
My mistake, A must be infinite.
$endgroup$
– Ángela Flores
Dec 5 '18 at 20:40




$begingroup$
My mistake, A must be infinite.
$endgroup$
– Ángela Flores
Dec 5 '18 at 20:40












$begingroup$
@AlexR. $l^2$ refers to the space of sequences $(a_1, a_2, ldots)$ such that $sum_n a_n^2 < infty$.
$endgroup$
– angryavian
Dec 5 '18 at 20:47




$begingroup$
@AlexR. $l^2$ refers to the space of sequences $(a_1, a_2, ldots)$ such that $sum_n a_n^2 < infty$.
$endgroup$
– angryavian
Dec 5 '18 at 20:47












$begingroup$
Nitpick: You might want to add $x ne y$ in the formulation of your question. Otherwise, it is not possible (obviously).
$endgroup$
– gerw
Dec 6 '18 at 7:52




$begingroup$
Nitpick: You might want to add $x ne y$ in the formulation of your question. Otherwise, it is not possible (obviously).
$endgroup$
– gerw
Dec 6 '18 at 7:52












$begingroup$
An interesting spinoff: How large can $$inf_{x,y in A, x ne y} |x-y|$$ be? Can this be larger than $sqrt{2}$?
$endgroup$
– gerw
Dec 6 '18 at 8:04






$begingroup$
An interesting spinoff: How large can $$inf_{x,y in A, x ne y} |x-y|$$ be? Can this be larger than $sqrt{2}$?
$endgroup$
– gerw
Dec 6 '18 at 8:04












1 Answer
1






active

oldest

votes


















2












$begingroup$

Recall that $ell^2$ is a Hilbert space with the natural inner product. Note that:
$$||x-y||^2=||x||^2+||y||^2-2langle x,yrangle,$$
in particular, if $x,yin S$ ($S$ the set of points with norm equal to $1$)
$$||x-y||^2=2(1-langle x,yrangle).$$



So it is enough to find an infinite family $B$ in $ell^2$ such that their inner product is $langle x,yrangle<0$ for all $x,yin B$, $xneq y$. Since you can consider
$$A=leftlbrace x/||x||;:;xin B, xneq 0rightrbrace,$$
and we have for $x/||x||$, $y/||y||$, $x,yin Bsetminus{0}$ the following
begin{align} ||x-y||^2 & = 2left( 1- frac{1}{||x||cdot||y||}langle x,yrangleright) \ & > 2.end{align}



Let $f:mathbb{N}tomathbb{N}:nmapsto n!$. Clearly $f(n)^2-nf(n-1)^2>0$. Now, let $(e_n)_n$ the standard basis of $ell^2$, and we define
$$u_n=f(n)e_n-sum_{k=1}^{n-1}f(k)e_k.$$
The set $B$ will be the set of the $u_n$'s. Let's check that for every pair of distinct natural numbers $n,m$ it is satisfied that $langle u_n,u_mrangle<0$. In fact
begin{align}langle u_n,u_mrangle & = leftlangle f(n)e_n-sum_{k=1}^{n-1}f(k)e_k,f(m)e_n-sum_{j=1}^{m-1}f(j)e_jrightrangle \ & = leftlangle f(n)e_n, f(m)e_mrightrangle-leftlangle f(n)e_n,sum_{j=1}^{m-1}f(j)e_jrightrangle \ & + leftlanglesum_{k=1}^{n-1}f(k)e_k,sum_{j=1}^{m-1}f(j)e_jrightrangle-leftlangle sum_{k=1}^{n-1}f(k)e_k, f(m)e_mrightrangle \ & = sum_{k=1}^{n-1}sum_{j=1}^{m-1}f(k)f(j)langle e_k,e_jrangle-f(m)^2 \ & = sum_{j=1}^{m-1}f(k)^2-f(m)^2 \ &leq mf(m-1)-f(m)^2 \ & < 0.end{align}






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    2












    $begingroup$

    Recall that $ell^2$ is a Hilbert space with the natural inner product. Note that:
    $$||x-y||^2=||x||^2+||y||^2-2langle x,yrangle,$$
    in particular, if $x,yin S$ ($S$ the set of points with norm equal to $1$)
    $$||x-y||^2=2(1-langle x,yrangle).$$



    So it is enough to find an infinite family $B$ in $ell^2$ such that their inner product is $langle x,yrangle<0$ for all $x,yin B$, $xneq y$. Since you can consider
    $$A=leftlbrace x/||x||;:;xin B, xneq 0rightrbrace,$$
    and we have for $x/||x||$, $y/||y||$, $x,yin Bsetminus{0}$ the following
    begin{align} ||x-y||^2 & = 2left( 1- frac{1}{||x||cdot||y||}langle x,yrangleright) \ & > 2.end{align}



    Let $f:mathbb{N}tomathbb{N}:nmapsto n!$. Clearly $f(n)^2-nf(n-1)^2>0$. Now, let $(e_n)_n$ the standard basis of $ell^2$, and we define
    $$u_n=f(n)e_n-sum_{k=1}^{n-1}f(k)e_k.$$
    The set $B$ will be the set of the $u_n$'s. Let's check that for every pair of distinct natural numbers $n,m$ it is satisfied that $langle u_n,u_mrangle<0$. In fact
    begin{align}langle u_n,u_mrangle & = leftlangle f(n)e_n-sum_{k=1}^{n-1}f(k)e_k,f(m)e_n-sum_{j=1}^{m-1}f(j)e_jrightrangle \ & = leftlangle f(n)e_n, f(m)e_mrightrangle-leftlangle f(n)e_n,sum_{j=1}^{m-1}f(j)e_jrightrangle \ & + leftlanglesum_{k=1}^{n-1}f(k)e_k,sum_{j=1}^{m-1}f(j)e_jrightrangle-leftlangle sum_{k=1}^{n-1}f(k)e_k, f(m)e_mrightrangle \ & = sum_{k=1}^{n-1}sum_{j=1}^{m-1}f(k)f(j)langle e_k,e_jrangle-f(m)^2 \ & = sum_{j=1}^{m-1}f(k)^2-f(m)^2 \ &leq mf(m-1)-f(m)^2 \ & < 0.end{align}






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Recall that $ell^2$ is a Hilbert space with the natural inner product. Note that:
      $$||x-y||^2=||x||^2+||y||^2-2langle x,yrangle,$$
      in particular, if $x,yin S$ ($S$ the set of points with norm equal to $1$)
      $$||x-y||^2=2(1-langle x,yrangle).$$



      So it is enough to find an infinite family $B$ in $ell^2$ such that their inner product is $langle x,yrangle<0$ for all $x,yin B$, $xneq y$. Since you can consider
      $$A=leftlbrace x/||x||;:;xin B, xneq 0rightrbrace,$$
      and we have for $x/||x||$, $y/||y||$, $x,yin Bsetminus{0}$ the following
      begin{align} ||x-y||^2 & = 2left( 1- frac{1}{||x||cdot||y||}langle x,yrangleright) \ & > 2.end{align}



      Let $f:mathbb{N}tomathbb{N}:nmapsto n!$. Clearly $f(n)^2-nf(n-1)^2>0$. Now, let $(e_n)_n$ the standard basis of $ell^2$, and we define
      $$u_n=f(n)e_n-sum_{k=1}^{n-1}f(k)e_k.$$
      The set $B$ will be the set of the $u_n$'s. Let's check that for every pair of distinct natural numbers $n,m$ it is satisfied that $langle u_n,u_mrangle<0$. In fact
      begin{align}langle u_n,u_mrangle & = leftlangle f(n)e_n-sum_{k=1}^{n-1}f(k)e_k,f(m)e_n-sum_{j=1}^{m-1}f(j)e_jrightrangle \ & = leftlangle f(n)e_n, f(m)e_mrightrangle-leftlangle f(n)e_n,sum_{j=1}^{m-1}f(j)e_jrightrangle \ & + leftlanglesum_{k=1}^{n-1}f(k)e_k,sum_{j=1}^{m-1}f(j)e_jrightrangle-leftlangle sum_{k=1}^{n-1}f(k)e_k, f(m)e_mrightrangle \ & = sum_{k=1}^{n-1}sum_{j=1}^{m-1}f(k)f(j)langle e_k,e_jrangle-f(m)^2 \ & = sum_{j=1}^{m-1}f(k)^2-f(m)^2 \ &leq mf(m-1)-f(m)^2 \ & < 0.end{align}






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Recall that $ell^2$ is a Hilbert space with the natural inner product. Note that:
        $$||x-y||^2=||x||^2+||y||^2-2langle x,yrangle,$$
        in particular, if $x,yin S$ ($S$ the set of points with norm equal to $1$)
        $$||x-y||^2=2(1-langle x,yrangle).$$



        So it is enough to find an infinite family $B$ in $ell^2$ such that their inner product is $langle x,yrangle<0$ for all $x,yin B$, $xneq y$. Since you can consider
        $$A=leftlbrace x/||x||;:;xin B, xneq 0rightrbrace,$$
        and we have for $x/||x||$, $y/||y||$, $x,yin Bsetminus{0}$ the following
        begin{align} ||x-y||^2 & = 2left( 1- frac{1}{||x||cdot||y||}langle x,yrangleright) \ & > 2.end{align}



        Let $f:mathbb{N}tomathbb{N}:nmapsto n!$. Clearly $f(n)^2-nf(n-1)^2>0$. Now, let $(e_n)_n$ the standard basis of $ell^2$, and we define
        $$u_n=f(n)e_n-sum_{k=1}^{n-1}f(k)e_k.$$
        The set $B$ will be the set of the $u_n$'s. Let's check that for every pair of distinct natural numbers $n,m$ it is satisfied that $langle u_n,u_mrangle<0$. In fact
        begin{align}langle u_n,u_mrangle & = leftlangle f(n)e_n-sum_{k=1}^{n-1}f(k)e_k,f(m)e_n-sum_{j=1}^{m-1}f(j)e_jrightrangle \ & = leftlangle f(n)e_n, f(m)e_mrightrangle-leftlangle f(n)e_n,sum_{j=1}^{m-1}f(j)e_jrightrangle \ & + leftlanglesum_{k=1}^{n-1}f(k)e_k,sum_{j=1}^{m-1}f(j)e_jrightrangle-leftlangle sum_{k=1}^{n-1}f(k)e_k, f(m)e_mrightrangle \ & = sum_{k=1}^{n-1}sum_{j=1}^{m-1}f(k)f(j)langle e_k,e_jrangle-f(m)^2 \ & = sum_{j=1}^{m-1}f(k)^2-f(m)^2 \ &leq mf(m-1)-f(m)^2 \ & < 0.end{align}






        share|cite|improve this answer









        $endgroup$



        Recall that $ell^2$ is a Hilbert space with the natural inner product. Note that:
        $$||x-y||^2=||x||^2+||y||^2-2langle x,yrangle,$$
        in particular, if $x,yin S$ ($S$ the set of points with norm equal to $1$)
        $$||x-y||^2=2(1-langle x,yrangle).$$



        So it is enough to find an infinite family $B$ in $ell^2$ such that their inner product is $langle x,yrangle<0$ for all $x,yin B$, $xneq y$. Since you can consider
        $$A=leftlbrace x/||x||;:;xin B, xneq 0rightrbrace,$$
        and we have for $x/||x||$, $y/||y||$, $x,yin Bsetminus{0}$ the following
        begin{align} ||x-y||^2 & = 2left( 1- frac{1}{||x||cdot||y||}langle x,yrangleright) \ & > 2.end{align}



        Let $f:mathbb{N}tomathbb{N}:nmapsto n!$. Clearly $f(n)^2-nf(n-1)^2>0$. Now, let $(e_n)_n$ the standard basis of $ell^2$, and we define
        $$u_n=f(n)e_n-sum_{k=1}^{n-1}f(k)e_k.$$
        The set $B$ will be the set of the $u_n$'s. Let's check that for every pair of distinct natural numbers $n,m$ it is satisfied that $langle u_n,u_mrangle<0$. In fact
        begin{align}langle u_n,u_mrangle & = leftlangle f(n)e_n-sum_{k=1}^{n-1}f(k)e_k,f(m)e_n-sum_{j=1}^{m-1}f(j)e_jrightrangle \ & = leftlangle f(n)e_n, f(m)e_mrightrangle-leftlangle f(n)e_n,sum_{j=1}^{m-1}f(j)e_jrightrangle \ & + leftlanglesum_{k=1}^{n-1}f(k)e_k,sum_{j=1}^{m-1}f(j)e_jrightrangle-leftlangle sum_{k=1}^{n-1}f(k)e_k, f(m)e_mrightrangle \ & = sum_{k=1}^{n-1}sum_{j=1}^{m-1}f(k)f(j)langle e_k,e_jrangle-f(m)^2 \ & = sum_{j=1}^{m-1}f(k)^2-f(m)^2 \ &leq mf(m-1)-f(m)^2 \ & < 0.end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 23:24









        José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda

        802110




        802110






























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            Mont Emei

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            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
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